# Question about reversible heat engine efficiency

I have a question regarding heat engines that cropped up whilst I was doing a practice question. I will summarise the results I obtained for the previous parts of the question so as to save your time. The highlighted parts of the image are where I am having some issues.

I confused because: -I thought all reversible heat engines operate exactly at the carnot efficiency $\frac{T_H - T_C}{T_H}$ -I thought that in theory, the engine cycle below is reversible and so should operate at this efficiency -From my workings, I found this not to be the case and I also found that there is an increase in entropy in the universe after each engine cycle (a reversible carnot engine would have no net change in entropy)

So my question is - are my workings wrong or can my results be explained as I misunderstand something? attempt at a solution (Important Results)

$$\gamma = \frac{5}{3}$$ $$T_2 = 2T_1$$ $$T_3 = \frac{p_3}{p_1}2T_1$$ $$T_3 = \left(\frac{1}{2}\right)^{2/3}T_1 = 0.63T_1$$ $$W_{31} = p_{1}V_{1}\left( \frac{2^{1-\gamma} - 1}{1-\gamma}\right) = 0.56p_{1}V_{1}$$ $$Q_{23} = \frac{3}{2}\left( (1/2)^{2/3} - 2\right)p_{1}V_{1} = -2.06p_{1}V_{1}$$ $$Efficiency = \frac{W_{net}}{Q_{Hot}} = \frac{W_{net}}{Q_{Cold} + W_{net}}$$ $W_{net} =$area enclosed by loop $= p_{1}V_{1}\left(1 + \frac{1-2^{1-\gamma}}{1-\gamma}\right)$ $$Q_{Cold} = Q_{23}$$ $$Efficiency = 0.18$$ $$\Delta S_{HotRes} = \frac{Q_{in}}{T_H} = \frac{-p_{1}V_{1}}{T_{1}}$$ $$\Delta S_{universe} = \Delta S_{HotRes} + \Delta S_{ColdRes}$$ $$\Delta S_{ColdRes} = -\frac{Q_{23}}{0.5T_1} = 3(2-(1/2)^{2/3})\frac{p_{1}V_{1}}{T_{1}}$$ $$\Delta S_{universe} = 3.11\frac{p_{1}V_{1}}{T_{1}}$$ $$Carnot efficiency = \frac{T_{H} - T_{C}}{T_{H}} = 0.8$$

So in short - why does the efficiency I've calculated differ from the Carnot efficiency and why do I find there to be a net increase in entropy.

• Why should all heat engines operate at the Carnot efficiency? The Otto cycle, for example, has a different efficiency... – Jon Custer Apr 11 '16 at 23:00
• @JonCuster: I think the OP is asking if all reversible engines have the same efficiency and why his doesn't, since he thinks it's reversible? – CuriousOne Apr 12 '16 at 0:11
• Your engine is not reversible because there is heat flow across finite temperature differences. For instance, the system and the hot reservoir (held fixed at $T_H$) are in contact during the isobaric expansion phase, and since the system and reservoir are at different temperatures during this process, this process is irreversible and hence the entire cycle is irreversible. – march Apr 12 '16 at 22:23
• Stage 2,3 is irreversible also, for the analogous reason. – Chet Miller Apr 15 '16 at 11:59

## 2 Answers

Entry of apparently accepted answer originally posted by march...

Your engine is not reversible because there is heat flow across finite temperature differences. For instance, the system and the hot reservoir (held fixed at THTH) are in contact during the isobaric expansion phase, and since the system and reservoir are at different temperatures during this process, this process is irreversible and hence the entire cycle is irreversible.

The most efficient possible heat engine is one running the Carnot Cycle.

All four 'steps' of the CC are reversible, so do not increase the entropy of the Universe (surroundings).

The work done is the area inside the loop, so a loop with 2 sides would also lose no entropy, but would convert no heat into motion, which is the purpose of a heat engine.

As an aside the most efficient 'machines' are large electrical transformers, which convert electrical energy to magnetic energy, and back to electrical energy - at a different voltage. The most efficient have an efficiency of over 99.9%.